Solution assignment 01 Optimizing f(x)

Determine the extreme of the parabola:

$y=x^2-7x+12$

We know that the graph of this function is an 'opens up' parabola and thus it has a minimum.
We differentiate the function and solve the equation:

$y'=2x-7=0$

The solution of this equation is:

$x=\displaystyle\frac{7}{2}$

and the coordinates of the minimum are:

$(\displaystyle\frac{7}{2},-\displaystyle\frac{1}{4})$